《2025年零失误分层训练高二数学选择性必修第三册黑吉辽内蒙古专版》

3. 【题型一】“杨辉三角”是中国古代数学文化的瑰宝之一，最早出现在中国南宋数学家杨辉于$1261$年所著的《详解九章算法》一书中，法国数学家帕斯卡$1654$年才发现这一规律. “杨辉三角”揭示了二项式系数在三角形数表中的一种几何排列规律，如图所示. 则下列关于“杨辉三角”的结论正确的是(C)


A.$C_3^3 + C_4^3 + C_5^3 + ·s + C_{10}^3 = 165$
B.在第$2022$行中，第$1011$个数最大
C.第$6$行的第$7$个数、第$7$行的第$7$个数及第$8$行的第$7$个数之和等于第$9$行的第$8$个数
D.第$34$行中第$15$个数与第$16$个数之比为$2:3$

答案： C[提示：由$C_{n}^{m - 1} + C_{n}^{m} = C_{n + 1}^{m}$可得$C_{10}^{2} + C_{10}^{3} + C_{10}^{4} + ·s + C_{10}^{10} = C_{11}^{3} + C_{11}^{4} + C_{11}^{5} + ·s + C_{11}^{10} = C_{12}^{4} + C_{12}^{5} + C_{12}^{6} + ·s + C_{12}^{10} = C_{13}^{5} = \frac{11×10×9}{3×2×1} = 165$，故A错误；第2022行中，第1011个数为$C_{2022}^{1011} \gt C_{2022}^{1010}$，故B错误；$C_{3}^{0} + C_{3}^{1} + C_{3}^{2} = C_{4}^{1} + C_{4}^{2} = C_{5}^{2}$，故C正确；第34行中第15个数与第16个数之比为$C_{34}^{14}:C_{34}^{15} = \frac{34×33×·s×21}{14×13×·s×1}:\frac{34×33×·s×20}{15×14×13×·s×1} = 15:20 = 3:4$，故D错误.]

4. 【题型二】设$(3x + \sqrt{x})^n$的展开式的各项系数之和为$M$，二项式系数之和为$N$，若$M - 17N = 480$，则展开式中含$x^3$项的系数为(D)

A.$40$
B.$30$
C.$20$
D.$15$

答案： D[提示：由$4^{n} - 17×2^{n} = 480$，得$n = 5$.$T_{k + 1} = C_{5}^{k}(3x)^{5 - k}(\sqrt{x})^{k} = 3^{5 - k}C_{5}^{k}x^{5 - \frac{k}{2}}$，令$5 - \frac{k}{2} = 3$，得$k = 4$.故展开式中含$x^{3}$项的系数为$3× C_{5}^{4} = 15$.]

5. 【题型三】$(2025·$黑龙江牡丹江一中高二下期中$)$ [多选] 已知$f(x) = (2x - 3)^n(n \in \mathbf{N}^*)$的展开式的二项式系数和为$512$，且$f(x) = a_0 + a_1(x - 1) + a_2(x - 1)^2 + ·s + a_n(x - 1)^n$，下列选项正确的是(BD)

A.$a_1 + a_2 + ·s + a_n = 1$
B.$a_1 + 2a_2 + 3a_3 + ·s + na_n = 18$
C.$a_2 = 144$
D.$|a_0| + |a_1| + ·s + |a_n| = 3^9$

答案： BD[提示：由已知有$2^{n} = C_{n}^{0} + C_{n}^{1} + ·s + C_{n}^{n} = 512$，故$n = 9$.$f(x) = (2x - 3)^{9}$，所以$(2x - 3)^{9} = a_{0} + a_{1}(x - 1) + a_{2}(x - 1)^{2} + ·s + a_{9}(x - 1)^{9}$.对于A，取$x = 1$得$a_{0} = (2 - 3)^{9} = -1$，取$x = 2$得$a_{0} + a_{1} + ·s + a_{9} = (4 - 3)^{9} = 1$，所以$a_{1} + a_{2} + ·s + a_{9} = 1 - (-1) = 2$，故A错误；对于B，对$(2x - 3)^{9} = a_{0} + a_{1}(x - 1) + a_{2}(x - 1)^{2} + ·s + a_{9}(x - 1)^{9}$求导得$18(2x - 3)^{8} = a_{1} + 2a_{2}(x - 1) + 3a_{3}(x - 1)^{2} + ·s + 9a_{9}(x - 1)^{8}$，取$x = 2$得$18 = a_{1} + 2a_{2} + 3a_{3} + ·s + 9a_{9}$，故B正确；对于C，在$(2x - 3)^{9} = a_{0} + a_{1}(x - 1) + a_{2}(x - 1)^{2} + ·s + a_{9}(x - 1)^{9}$中用$x + 1$替换$x$，得$(2x - 1)^{9} = a_{0} + a_{1}x + a_{2}x^{2} + ·s + a_{9}x^{9}$，所以$a_{k} = C_{9}^{k} · 2^{9 - k} · (-1)^{9 - k}(k = 0,1,·s,9)$，特别地，当$k = 2$时$a_{2} = C_{9}^{2} · 2^{7} · (-1)^{7} = -144$，故C错误；对于D，由$a_{k} = C_{9}^{k} · 2^{9 - k} · (-1)^{9 - k}(k = 0,1,·s,9)$有$|a_{0}| + |a_{1}| + ·s + |a_{9}| = -a_{0} + a_{1} - a_{2} + ·s - a_{9}$，在$(2x - 3)^{9} = a_{0} + a_{1}(x - 1) + a_{2}(x - 1)^{2} + ·s + a_{9}(x - 1)^{9}$中取$x = 0$得$-3^{9} = a_{0} - a_{1} + a_{2} - ·s - a_{9}$，所以$|a_{0}| + |a_{1}| + ·s + |a_{9}| = -a_{0} + a_{1} - a_{2} + ·s - a_{9} = 3^{9}$，故D正确.]

6. 【题型三】$(2025·$吉林长春外国语学校高二下期中$)$ [多选] 已知$f(x) = (2 - x)^8 = a_0 + a_1x + a_2x^2 + ·s + a_8x^8$，则下列描述不正确的是(ACD)

A.$a_1 + a_2 + ·s + a_8 = 1$
B.$f(-1)$除以$5$所得的余数是$1$
C.$|a_1| + |a_2| + |a_3| + ·s + |a_8| = 3^8$
D.$2a_2 + 3a_3 + ·s + 8a_8 = -8$

答案： ACD[提示：因为$f(x) = (2 - x)^{8} = a_{0} + a_{1}x + a_{2}x^{2} + ·s + a_{8}x^{8}$，所以令$x = 1$，可得$f(1) = a_{0} + a_{1} + a_{2} + ·s + a_{8} = 1$，再令$x = 0$，可得$a_{0} = 2^{8}$，所以$a_{1} + a_{2} + ·s + a_{8} = 1 - 2^{8} = -255$，故A错误.因为$|a_{0}| + |a_{1}| + |a_{2}| + ·s + |a_{8}|$可以看成$(2 + x)^{8}$的展开式各项系数之和，所以$|a_{0}| + |a_{1}| + |a_{2}| + ·s + |a_{8}| = 3^{8}$，所以$|a_{1}| + |a_{2}| + ·s + |a_{8}| = 3^{8} - 2^{8}$，故C错误.由题意，$f(-1) = 3^{8} = 9^{4} = (10 - 1)^{4} = C_{4}^{0} · 10^{4} - C_{4}^{1} · 10^{3} + C_{4}^{2} · 10^{2} - C_{4}^{3} · 10 + C_{4}^{4}$，显然，除了最后一项外，其余各项均能被5整除，所以$f(-1)$除以5所得的余数是1，故B正确.把函数$f(x)$两边同时对$x$求导数，可得$-8(2 - x)^{7} = a_{1} + 2a_{2}x + 3a_{3}x^{2} + ·s + 8a_{8}x^{7}$，再令$x = 1$，可得$a_{1} + 2a_{2} + 3a_{3} + ·s + 8a_{8} = -8$，令$x = 0$，可得$a_{1} = -8×2^{7}$，所以$2a_{2} + 3a_{3} + ·s + 8a_{8} = -8 - a_{1} = -8 + 8×2^{7}$，故D错误.]

7. 【题型二、三】$(2025·$重庆巴蜀中学高二下期中$)$ [多选] 已知$(3 - 2x)^6 = a_0 + a_1x + a_2x^2 + ·s + a_6x^6$，则(AD)

A.$a_0 = 729$
B.$a_3$是所有系数中的最大值
C.$a_0 + a_2 + a_4 + a_6 = \dfrac{5^6 - 1}{2}$
D.$2^6a_0 + 2^5a_1 + 2^4a_2 + ·s + 2a_5 + a_6 = 4096$

答案： AD[提示：对于A，令$x = 0$得$a_{0} = 3^{6} = 729$，故A正确；对于B，$a_{3}$是$x^{3}$的系数，而$C_{6}^{3} · 3^{3} · (-2x)^{3}$明显其系数小于零，不可能是所有系数中的最大值，故B错误；对于C，令$x = 1$得$a_{0} + a_{1} + a_{2} + ·s + a_{6} = 1$，令$x = -1$得$a_{0} - a_{1} + a_{2} + ·s + a_{6} = 5^{6}$，两式相加得$2(a_{0} + a_{2} + a_{4} + a_{6}) = 5^{6} + 1$，则$a_{0} + a_{2} + a_{4} + a_{6} = \frac{5^{6} + 1}{2}$，故C错误；对于D，令$x = \frac{1}{2}$得$a_{0} + a_{1}×\frac{1}{2} + a_{2}×(\frac{1}{2})^{2} + ·s + a_{6}×(\frac{1}{2})^{6} = 2^{6}$，等式两边同时乘$2^{6}$得$2^{6}a_{0} + 2^{5}a_{1} + 2^{4}a_{2} + ·s + 2a_{5} + a_{6} = 2^{12} = 4096$，故D正确.]

8. 【题型二】$(2025·$河北昌黎一中高二下期中$)$ [多选] 已知$\left(x^2 + \dfrac{1}{\sqrt{x}}\right)^n$的展开式中第$5$项与第$7$项的二项式系数相等，则下列说法正确的是(BCD)

A.展开式的各项系数之和为$4096$
B.展开式中含$x^{15}$项的系数为$45$
C.展开式中存在常数项
D.展开式中第$6$项的系数最大

答案： BCD[提示：由二项式的展开式中第5项与第7项的二项式系数相等，得$C_{n}^{4} = C_{n}^{6}$，解得$n = 10$，令$x = 1$，得$(1^{2} + \frac{1}{\sqrt{1}})^{10} = 1024$，即展开式的各项系数之和为1024，故A错误；由通项公式$T_{k + 1} = C_{10}^{k}x^{2(10 - k)}(\frac{1}{2})^{k}$，令$2(10 - k) - \frac{k}{2} = 15$，解得$k = 2$，所以展开式中含$x^{15}$项的系数为$C_{10}^{2} = 45$，故B正确；若展开式中存在常数项，令$2(10 - k) - \frac{k}{2} = 0$，解得$k = 8$，故C正确；由$n = 10$可知展开式共有11项，中间项的二项式系数最大，即第6项的系数最大，故D正确.]

9. 【题型三】若$(1 - 2x)^{2021} = a_0 + a_1x + a_2x^2 + ·s + a_{2021}x^{2021}(x \in \mathbf{R})$，则$\dfrac{a_1}{2} + \dfrac{a_2}{2^2} + ·s + \dfrac{a_{2021}}{2^{2021}}$的值为-1.

答案： -1[提示：因为$(1 - 2x)^{2021} = a_{0} + a_{1}x + a_{2}x^{2} + ·s + a_{2021}x^{2021}(x \in \mathbf{R})$，令$x = 0$可得$a_{0} = 1$；令$x = \frac{1}{2}$可得$a_{0} + \frac{a_{1}}{2} + \frac{a_{2}}{2^{2}} + ·s + \frac{a_{2021}}{2^{2021}} = (1 - 2×\frac{1}{2})^{2021} = 0$，故$\frac{a_{1}}{2} + \frac{a_{2}}{2^{2}} + ·s + \frac{a_{2021}}{2^{2021}} = 0 - a_{0} = -1$.]

1. 已知$(1 - x)^5 = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5$，则$(a_0 + a_2 + a_4)(a_1 + a_3 + a_5)$的值等于-256.

答案： -256[提示：令$x = 1$，得$a_{0} + a_{2} + a_{4} = -(a_{1} + a_{3} + a_{5}) = C_{5}^{0} + C_{5}^{2} + C_{5}^{4} = 16$，则$(a_{0} + a_{2} + a_{4}) · (a_{1} + a_{3} + a_{5}) = -256$.]

2. 已知$(2x - 1)^n$的展开式中，奇次项系数的和比偶次项系数的和小$3^8$，则$C_n^1 + C_n^2 + C_n^3 + ·s + C_n^n$的值为255.

答案： 255[提示：设$(2x - 1)^{n} = a_{0} + a_{1}x + a_{2}x^{2} + ·s + a_{n}x^{n}$，且奇次项的系数和为$A$，偶次项的系数和为$B$，则$A = a_{1} + a_{3} + a_{5} + ·s$，$B = a_{0} + a_{2} + a_{4} + a_{6} + ·s$，由已知得$B - A = 3^{8}$，令$x = -1$得$a_{0} - a_{1} + a_{2} - a_{3} + ·s + a_{n}(-1)^{n} = (-3)^{n}$，即$(a_{0} + a_{2} + a_{4} + a_{6} + ·s) - (a_{1} + a_{3} + a_{5} + a_{7} + ·s) = (-3)^{n}$，即$B - A = (-3)^{n}$，所以$(-3)^{n} = 3^{8} = (-3)^{8}$，所以$n = 8$，所以$C_{8}^{0} + C_{8}^{2} + C_{8}^{3} + ·s + C_{8}^{8} = 2^{n} - C_{8}^{0} = 2^{8} - 1 = 255$.]